• 每個線程用不同的線程函數(shù)
  例如：
  std::mutex mutex_;
  void func1()
  {
  ??? std::cout << "func1" << std::endl;
  ??? std::lock_guard<std::mutex> lck(mutex_);
  ??? std::this_thread::sleep_for(std::chrono::seconds(7));
  ??? std::cout << "func1 end" << std::endl;
  }
  
  void func2()
  {
  ??? std::cout << "func2" << std::endl;
  ??? std::lock_guard<std::mutex> lck(mutex_);
  ??? std::cout << "func2 coming" << std::endl;
  ??? std::this_thread::sleep_for(std::chrono::seconds(3));
  ??? std::cout << "func2 end" << std::endl;
  }
  
  int main()
  {
  ??? std::thread t1(func1);
  ??? std::this_thread::sleep_for(std::chrono::seconds(2));
  ??? std::thread t2(func2);
  ??? std::this_thread::sleep_for(std::chrono::seconds(15));
  }
  由于我們在定義t1線程后，休眠了2秒，則基本可以保證：func1函數(shù) 執(zhí)行到定義鎖且到了睡眠7秒的地方了，此時線程2才開始，
  由于線程1先獲取到鎖且睡眠7秒，因此：線程2必須等7秒后才能輸出 func2 coming

• 多個線程公用一個線程函數(shù)

  int counter = 0;
  void increase(int count) {
  ??? for (size_t i = 0; i < count; i++)
  ??? {
  ??????? /*std::lock_guard<std::mutex> lck(mutex_);*/
  ??????? std::this_thread::sleep_for(std::chrono::milliseconds(1));
  ??????? counter++;
  ??? }
  }

  測試：std::thread t1(increase, 5000);
  ??? std::thread t2(increase, 5000);
  ??? t1.join();
  ??? t2.join();
  ??? std::this_thread::sleep_for(std::chrono::seconds(5));
  ??? std::cout << "counter:" << counter << std::endl;

  結(jié)果：counter值并不是10000，說明了2個線程函數(shù)執(zhí)行時有對同counter變量修改存在問題，執(zhí)行3次結(jié)果都不同：

于是：通過枷鎖，來實現(xiàn)互斥訪問：
for (size_t i = 0; i < count; i++)
??? {
??????? std::lock_guard<std::mutex> lck(mutex_);
??????? std::this_thread::sleep_for(std::chrono::milliseconds(1));
??????? counter++;
??? }執(zhí)行3次結(jié)果如下，均為10000：

當然了，也可以在2個線程函數(shù)里執(zhí)行，枷鎖后效果一樣的。